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card trick without the trick...

The question could've been a bit more specific by stating "i then arrange the other four cards in a special way" as "i then 'order' the other four cards in a special way".

I figured that If i'm allowed to encode the card by "arranging" the four cards, I can keeo each card in either vertical or horizontal positions. This would allow me to encode 2^4 = 16 values. Ofcourse this is not enough so we move on a higher base system. I now keep a card horizontal or vertical in addition to keeping it above or below a reference line(or some such modification that allows us to use a higher base system e.g. keep the carc vertical, horizontal or tilted). This would encode 4^4=256 more than enough for the 52 value that we have...

Ofcourse its a minor issue... But I got the solution wrong :(

Anyways, thanks for the problems ... I'm addicted to puzzle solving.

~ Amol

Amol Deshmukh
Monday, December 2, 2002

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